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Mathematics 12 Online
OpenStudy (anonymous):

if P(E)=0.33, what are the odds against E?

OpenStudy (anonymous):

\[\text{Odds against $E$}=\frac{P(E')}{P(E)}=\frac{1-P(E)}{P(E)}\]

OpenStudy (anonymous):

why is it not just 1-.33 @SithsAndGiggles .

OpenStudy (anonymous):

Can you help me with my question? Please.

OpenStudy (anonymous):

1 - .33 is just the probability of "not E". OP is looking for odds.

OpenStudy (anonymous):

ok.

OpenStudy (anonymous):

it says _ to _ (type integers with no factors)

OpenStudy (anonymous):

@mvesling, do you mean as a decimal?

OpenStudy (anonymous):

out of 100 you lose 33, win 77

OpenStudy (anonymous):

no that is wrong sorry

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

out of 100 you lose 67, win 33 better

OpenStudy (anonymous):

odds against are losses to wins

OpenStudy (anonymous):

so 67 to 33?

OpenStudy (anonymous):

thx

OpenStudy (anonymous):

yes

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